GENERAL CHEMISTRY · GASES

Name the system.
Choose the law.

Learn all five official topics through decision ledgers, worked examples, retrieval prompts, and twenty original five-choice questions.

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5guided lessons

20practice questions

15study objectives

5choices per item

A repeatable gas-law routine

System. Constants. Units. Model.

01Name the system
Decide whether the prompt describes one sample, two states, or a mixture.
02Mark constants
Identify amount, temperature, pressure, or volume held fixed.
03Normalize units
Use kelvins, absolute pressure, and units compatible with the chosen R.
04Check the model
Confirm direction, scale, and whether the ideal approximation is reasonable.

Five prerequisite-aware lessons

Connect particles to equations.

Each lesson uses a semantic decision ledger, one fully worked example, and closed-note retrieval before multiple-choice practice.

LESSON 1 · 18 MIN

Study + retrieve

Kinetic molecular theory from particles to pressure

Use ideal-gas assumptions to explain pressure, temperature, molecular speed, and the conditions that expose real-gas behavior.

ESSENTIAL QUESTIONWhich microscopic change can explain the macroscopic observation without violating the model?

Particle-to-observation ledger

Ideal-gas assumptions and their observable consequences
Model statement	What it explains	Boundary
Continuous random motion	Particles repeatedly encounter the walls	Individual paths remain unpredictable
Elastic collisions	Total collision kinetic energy is conserved	Speed and direction can still change
Negligible particle volume	Most container volume is empty space	Fails progressively at high pressure
No modeled attractions	Particles separate without an energy penalty	Fails progressively at low temperature
Average KE ∝ kelvin T	Equal T means equal average KE	Equal KE does not mean equal speed

Mass, temperature, and root-mean-square speed
Comparison	Relationship	Conclusion
Same gas, higher T	uᵣₘₛ ∝ √T	Faster molecular motion
Same T, lighter gas	uᵣₘₛ ∝ 1/√M	Higher root-mean-square speed
Same T, any ideal gas	Average KE is equal	Mass changes speed, not average KE

STUDY TABLE · CAPTION INCLUDED

Start from the postulates

Ideal-gas particles move continuously, are negligibly small relative to their separations, exert no modeled attractions or repulsions, and collide elastically. Pressure comes from collisions with the container walls.

Model assumptions are not literal particle properties
Elastic means total kinetic energy is conserved in collisions

Separate energy from speed

Average translational kinetic energy depends only on kelvin temperature. At the same temperature, light and heavy gases have equal average kinetic energy, but lighter particles move faster on average because the same energy is carried by less mass.

Same T → same average KE
uᵣₘₛ ∝ √(T/M)

Know when ideality weakens

Lower pressure and higher temperature generally improve the ideal approximation. High pressure makes finite particle volume important; low temperature makes attractive forces more consequential and brings condensation closer.

Low P · high T → more ideal
Opposing P and T changes may be indeterminate

Worked example

At the same temperature, compare the root-mean-square speeds of gases with molar masses 4.00 and 36.0 g/mol.

1
At equal temperature, uᵣₘₛ is inversely proportional to the square root of molar mass.
2
Form the light-to-heavy speed ratio: √(36.0/4.00).
3
Evaluate √9.00 = 3.00.

ConclusionThe 4.00 g/mol gas moves three times as fast by the root-mean-square measure, while both gases have the same average kinetic energy.

Close the notes first

Retrieve the model.

01What microscopic event creates gas pressure?

Particle collisions with the container walls.

The frequency and momentum transfer of wall collisions produce the macroscopic pressure.

02At one temperature, do helium and xenon have the same average kinetic energy or the same molecular speed?

The same average kinetic energy, not the same speed.

The lighter helium particles move faster on average to carry the same average kinetic energy.

03Which paired change makes ideal behavior more likely?

Lower pressure and higher temperature.

Particles are farther apart, so finite volume and attractions matter less.

LESSON 2 · 17 MIN

Study + retrieve

Dalton’s law with mixture and water-vapor ledgers

Compute total pressure, mole-fraction partial pressure, and dry-gas pressure while preserving mixture and experimental boundaries.

ESSENTIAL QUESTIONWhich gases are actually contributing to the measured pressure?

Mixture pressure ledger

Choosing the Dalton relationship from the available evidence
Known information	Relationship	Required check
Component pressures	Ptotal = ΣPᵢ	Use compatible pressure units
Component and total moles	Xᵢ = nᵢ/ntotal	Mole fractions sum to 1
Mole fraction and total pressure	Pᵢ = XᵢPtotal	0 ≤ Pᵢ ≤ Ptotal
Gas collected over water	Pgas = Ptotal − Pwater	Use vapor pressure at the stated T

Worked mixture: 2.00 mol O₂ and 6.00 mol N₂ at 4.00 atm
Component	Mole fraction	Partial pressure
O₂	2.00/8.00 = 0.250	1.00 atm
N₂	6.00/8.00 = 0.750	3.00 atm
Total	1.000	4.00 atm

STUDY TABLE · CAPTION INCLUDED

Inventory every gas

For a nonreacting ideal mixture, total pressure is the sum of the component partial pressures. Put every pressure in compatible units before adding or subtracting and make sure no partial pressure exceeds the total.

Ptotal = ΣPᵢ
Partial pressures are nonnegative

Translate amount into pressure

At common temperature and volume, a component’s share of pressure equals its mole fraction: Xᵢ = nᵢ/ntotal and Pᵢ = XᵢPtotal. Mass fractions must first be converted to mole fractions.

ΣXᵢ = 1
Use moles, not grams

Remove water vapor correctly

Gas collected over water is a mixture of the target gas and water vapor. When the stated measurement is the total pressure, subtract the supplied water-vapor pressure to isolate the dry gas.

Pgas = Ptotal − Pwater
Water-vapor pressure depends on temperature

Worked example

A mixture contains 2.00 mol O₂ and 6.00 mol N₂ at a total pressure of 4.00 atm. Find the O₂ partial pressure.

1
Find total amount: 2.00 + 6.00 = 8.00 mol.
2
Calculate the O₂ mole fraction: 2.00/8.00 = 0.250.
3
Multiply by total pressure: 0.250(4.00 atm) = 1.00 atm.

ConclusionO₂ contributes 1.00 atm and N₂ contributes the remaining 3.00 atm.

Close the notes first

Retrieve the model.

01Three partial pressures are 0.20, 0.30, and 0.40 atm. What is Ptotal?

0.90 atm.

Dalton’s law adds all compatible partial pressures.

02A gas has mole fraction 0.30 at 2.00 atm total pressure. What is its partial pressure?

0.60 atm.

Pᵢ = XᵢPtotal.

03Why is water-vapor pressure subtracted from a gas-over-water total?

Water vapor is already one component of the measured mixture.

Subtracting it leaves only the target gas’s partial pressure.

LESSON 3 · 16 MIN

Study + retrieve

Boyle’s law with absolute pressure and graph checks

Recognize constant-temperature pressure–volume problems, calculate a second state, and identify the inverse graph without using gauge pressure as absolute pressure.

ESSENTIAL QUESTIONDid only pressure and volume change for the same gas sample?

Constant-temperature state ledger

Boyle-law variables and direction checks
State	Absolute pressure	Volume	Product
Initial	P₁	V₁	P₁V₁
Final	P₂	V₂	P₂V₂
Fixed n and T	P₁V₁ = P₂V₂		Constant

Graph and pressure interpretation
Evidence	Correct interpretation	Common trap
P versus V	Decreasing hyperbola	Drawing a straight descending line
P versus 1/V	Increasing line through the origin	Reversing the slope
Gauge pressure	Add supplied ambient pressure	Using gauge pressure as absolute

STUDY TABLE · CAPTION INCLUDED

Confirm the fixed variables

Boyle’s law applies to a fixed amount of gas at constant temperature. Under those conditions, pressure and volume change inversely and their product remains constant.

n and T fixed
P₁V₁ = P₂V₂

Use absolute pressure

Gas-law ratios require absolute pressure. If a pressure is reported relative to the surroundings, combine the gauge reading with the supplied ambient pressure before using it.

Pabsolute = Pgauge + Pambient
Direction check: smaller V → larger P

Read the graph correctly

P versus V is a decreasing hyperbola, not a straight descending line. P versus 1/V and 1/P versus V are increasing straight lines through the origin for the ideal relationship.

Direct axes → hyperbola
Reciprocal axis → line

Worked example

A gas occupies 2.50 L at 1.20 atm and is compressed to 1.50 L at constant temperature. Find its final pressure.

1
Confirm the same amount and constant temperature, so Boyle’s law applies.
2
Rearrange: P₂ = P₁V₁/V₂.
3
Calculate 1.20 atm × 2.50 L / 1.50 L = 2.00 atm.

ConclusionThe pressure rises to 2.00 atm, consistent with the smaller final volume.

Close the notes first

Retrieve the model.

01If absolute pressure doubles at constant n and T, what happens to volume?

It halves.

PV is constant under Boyle conditions.

02A gauge reads 0.40 atm while ambient pressure is 1.00 atm. What pressure enters Boyle’s law?

1.40 atm absolute.

Gauge pressure is measured relative to the surroundings.

03What shape is a P-versus-V Boyle plot?

A decreasing hyperbola.

P is proportional to 1/V, not directly to V.

LESSON 4 · 16 MIN

Study + retrieve

Charles’s law on an absolute temperature scale

Recognize constant-pressure volume–temperature problems, calculate a second state in kelvins, and interpret absolute zero as a model boundary.

ESSENTIAL QUESTIONIs temperature being used as an absolute ratio rather than a Celsius difference?

Absolute-temperature state ledger

Charles-law calculation sequence
Step	Initial state	Final state	Check
Convert	T₁(°C) + 273.15	T₂(°C) + 273.15	Both temperatures are positive K
Relate	V₁/T₁	V₂/T₂	n and P are fixed
Predict	Higher K ↔ larger V		Direction matches result

Temperature-scale boundaries
Statement	Use	Do not infer
V ∝ T	Kelvin ratios	Celsius ratios
Line extrapolates to 0 K	Defines ideal-model intercept	A real gas reaches zero volume
Cooling toward condensation	Expect growing nonideality	Ideal behavior at every temperature

STUDY TABLE · CAPTION INCLUDED

Confirm constant pressure and amount

For a fixed amount of gas at constant pressure, volume is directly proportional to absolute temperature. Heating requires expansion to keep wall-collision pressure constant.

n and P fixed
V/T = constant

Convert before forming a ratio

Celsius has an offset zero and cannot be used directly in a proportional gas-law ratio. Convert each state separately with T(K) = T(°C) + 273.15.

Convert both temperatures
Warmer K → larger V

Treat absolute zero as a limit

An ideal V-versus-K line extrapolates to the origin, corresponding to −273.15 °C. A real gas condenses and becomes nonideal before a constant-pressure path could reach zero volume.

Ideal-model extrapolation
Not a physical zero-volume claim

Worked example

A gas occupies 1.50 L at 27.0 °C. What volume would it occupy at −23.0 °C if pressure and amount stay constant?

1
Convert the two temperatures: 300.15 K and 250.15 K.
2
Use V₂ = V₁T₂/T₁.
3
Calculate 1.50 L × 250.15/300.15 = 1.25 L.

ConclusionThe gas contracts to about 1.25 L; the direction agrees with the lower absolute temperature.

Close the notes first

Retrieve the model.

01Why can 20 °C and 40 °C not be used as a 1:2 gas-law temperature ratio?

Celsius does not begin at absolute zero.

The corresponding values are about 293 K and 313 K, not a doubling.

02At constant pressure and amount, what happens to volume when kelvin temperature rises?

Volume rises in direct proportion.

Expansion offsets the increased particle motion so pressure remains fixed.

03Does the Charles-law intercept prove a real gas reaches zero volume?

No.

The ideal line is extrapolated beyond the region where a real gas remains gaseous and ideal.

LESSON 5 · 20 MIN

Study + retrieve

Ideal gas law with unit and model selection

Use PV = nRT, combined-state reasoning, and gas density relationships with compatible units, absolute scales, and an explicit ideality check.

ESSENTIAL QUESTIONWhich variables, units, and assumptions belong in this gas model?

Gas-model and unit ledger

Selecting the smallest valid gas relationship
System	Fixed information	Direct model
One state; amount required	Known P, V, T	PV = nRT
Two states; same sample	n fixed; P, V, T change	P₁V₁/T₁ = P₂V₂/T₂
Two states; constant T	n and T fixed	P₁V₁ = P₂V₂
Two states; constant P	n and P fixed	V₁/T₁ = V₂/T₂
Nonreacting gas mixture	Common container	Ptotal = ΣPᵢ

Common gas constants and compatible pressure–volume units
Gas constant	Use with	Still required
0.082057 L·atm·mol⁻¹·K⁻¹	Liters and atmospheres	Absolute P and kelvin T
8.314 L·kPa·mol⁻¹·K⁻¹	Liters and kilopascals	Absolute P and kelvin T

Derived formsM = dRT/P · d = PM/RT

STUDY TABLE · CAPTION INCLUDED

Match every unit to R

PV = nRT relates one gas state. Pressure must be absolute, temperature must be in kelvins, and the pressure–volume units must match the chosen gas constant.

0.082057 L·atm·mol⁻¹·K⁻¹
8.314 L·kPa·mol⁻¹·K⁻¹

Select the smallest valid model

Use Boyle or Charles when their omitted variables are fixed. Use the combined gas law for a fixed amount when P, V, and T all change, Dalton for a mixture, and PV = nRT when amount is needed or changes.

Name constants before equations
P₁V₁/T₁ = P₂V₂/T₂ when n is fixed

Connect amount, mass, and density

Substitute n = m/M and d = m/V into PV = nRT to obtain M = dRT/P or d = PM/RT. Do not insert a standard molar volume unless the stated conditions match that convention.

Track g/L with L-based R
High P or low T weakens ideality

Worked example

What volume does 0.500 mol of an ideal gas occupy at 300 K and 1.00 atm?

1
Choose R = 0.082057 L·atm·mol⁻¹·K⁻¹ to match pressure in atm.
2
Rearrange PV = nRT to V = nRT/P.
3
Calculate 0.500 × 0.082057 × 300 / 1.00 = 12.3 L.

ConclusionThe gas occupies 12.3 L to three significant figures under the ideal-gas assumption.

Close the notes first

Retrieve the model.

01Which R value pairs directly with kPa and liters?

8.314 L·kPa·mol⁻¹·K⁻¹.

Its pressure–volume units cancel kPa·L in PV.

02Which law compares two states of one fixed gas sample when P, V, and T all change?

The combined gas law.

The amount cancels between the two ideal-gas states.

03What gas-law expression gives molar mass from density?

M = dRT/P.

It follows from substituting n = m/M and d = m/V into PV = nRT.

All twenty Gases problems

Select the valid model.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

20 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions have not been calibrated to the official score scale. Use explanations to repair the specific reasoning step, then use balanced practice to switch among all five topics.

DAT TRAIN does not claim topic quotas because the official manual does not publish them.